I am working on a MonteCarlo simulation model and part of it is to calculate the following formula:
X = Sqr(1-p)Y + Sqr(p)Z,
Where:
Y and Z are randomly obtained values based (idiosyncratic and systematic factors, respectviely) on a standard normal (inv.) distribution, calculated as:
Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
p represents a correlation factor.
My aim is to square root a recalled formula, however when I try the following (inserting the first Sqr), it does not work and gives an error:
Matrix (n, sims) = (R * Sqr(Application.WorksheetFunction.NormInv(Rnd(), mean, sd))) + (Sqr(1 - R) * RandomS(s, x))
where:
R: Correlation factor
RandomS(s,x): generated matrix with Z values.
I don't want to go into too much details about the background and other variables, as the only problem I am getting is with Square Rooting the equation.
Error message I recieve reads:
Run-time error '5':
Invalid procedure call or argument
When I click debug it takes me to the formula, therefore there must be something wrong with the syntax.
Can you help with directly squaring the formula?
Thank you!
Andrew
Square root is simply Sqr.
It works fine in Excel VBA, so for example:
MsgBox Sqr(144)
...returns 12.
Just don't confuse it with the syntax for a worksheet function with is SQRT.
If you're still having an issue with your formula, tit must be with something other than the Square Root function, and I'd suggest you check the values of your variable, and make sure they are properly declared (preferably with Option Explicit at the top of the module).
Also make sure that you're passing Sqr a positive number.
Documentation: Sqr Function
I'm not a math major, but with your formula:
X = Sqr(1-p)Y + Sqr(p)Z,
...you specified how Y and Z are calculated, so calculate them separately to keep it simple:
Dim X as Double, Y as Double, Z as Double
Y = Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
Z = Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
Assuming the comma is not supposed to be in the formula, and having no idea what p is, your final code to calculate X is:
X = Sqr(1-p) * Y + Sqr(p) * Z
Related
I've got this equation from mathematical model to know the thermal behavior of a battery.
dTsdt = Ts * a+ Ta * b + dTadt * c + d
However, i can't get to solve it due to the nested derivatives.
I need to solve the equation for Ts and Ta.
I tried to define it as follows, but python does not like it and several eŕrors show up.
Im using scipy.integrate and the solver ODEint
Since the model takes data from vectors, it has to be solved for every time step and record the output accordingly.
I also tried assinging the derivatives to a variable v1,v2, and then put everything in an equation without derivatives like the second approach shown as follows.
def Tmodel(z,t,a,b,c,d):
Ts,Ta= z
dTsdt = Ts*a+ Ta*b + dTadt*c+ d
dzdt=[dTsdt]
return dzdt
z0=[0,0]
# solve ODE
for i in range(0,n-1):
tspan = [t[i],t[i+1]]
# solve for next step
z = odeint(Tmodel,z0,tspan,arg=(a[i],b[i],c[i],d[i],))
# store solution for plotting
Ts[i] = z[1][0]
Ta[i] = z[1][1]
# next initial condition
z0 = z[1]
def Tmodel(z,t,a,b,c,d):
Ts,v1,Ta,v2= z
# v1= dTsdt
# v2= dTadt
v1 = Ts*a+ Ta*b + v2*c+ d
dzdt=[v1,v2]
return dzdt
That did not work either.I believe there might be a solver capable of solving that equation or the equation must be decouple in a way and solve accordingly.
Any advice on how to solve such eqtn with python would be appreciate it.
Best regards,
MM
Your difficulty seems to be that you are given Ta in a form with no easy derivative, so you do not know where to take it from. One solution is to avoid this derivative completely and solve the system for y=Ts-c*Ta. Substitute Ts=y+c*Ta in the right side to get
dy/dt = y*a + Ta*(b+c*a) + d
Of course, this requires then a post-processing step Ts=y+c*Ta to get to the requested variable.
If Ta is given as function table, use an interpolation function to get values at any odd time t that is demanded by the ODE solver.
Ta_func = interp1d(Ta_times,Ta_values)
def Tmodel(y,t,a,b,c,d):
Ta= Ta_func(t)
dydt = y*a+ Ta*(b+c*a) + d
return dydt
y[0] = Ts0-c*Ta_func(t[0])
for i in range(len(t)-1):
y[i+1] = odeint(Tmodel,y[i],t[i:i+2],arg=(a[i],b[i],c[i],d[i],))[-1,0]
Ts = y + c*Ta_func(t)
I'm attempting to solve the differential equation:
m(t) = M(x)x'' + C(x, x') + B x'
where x and x' are vectors with 2 entries representing the angles and angular velocity in a dynamical system. M(x) is a 2x2 matrix that is a function of the components of theta, C is a 2x1 vector that is a function of theta and theta' and B is a 2x2 matrix of constants. m(t) is a 2*1001 array containing the torques applied to each of the two joints at the 1001 time steps and I would like to calculate the evolution of the angles as a function of those 1001 time steps.
I've transformed it to standard form such that :
x'' = M(x)^-1 (m(t) - C(x, x') - B x')
Then substituting y_1 = x and y_2 = x' gives the first order linear system of equations:
y_2 = y_1'
y_2' = M(y_1)^-1 (m(t) - C(y_1, y_2) - B y_2)
(I've used theta and phi in my code for x and y)
def joint_angles(theta_array, t, torques, B):
phi_1 = np.array([theta_array[0], theta_array[1]])
phi_2 = np.array([theta_array[2], theta_array[3]])
def M_func(phi):
M = np.array([[a_1+2.*a_2*np.cos(phi[1]), a_3+a_2*np.cos(phi[1])],[a_3+a_2*np.cos(phi[1]), a_3]])
return np.linalg.inv(M)
def C_func(phi, phi_dot):
return a_2 * np.sin(phi[1]) * np.array([-phi_dot[1] * (2. * phi_dot[0] + phi_dot[1]), phi_dot[0]**2])
dphi_2dt = M_func(phi_1) # (torques[:, t] - C_func(phi_1, phi_2) - B # phi_2)
return dphi_2dt, phi_2
t = np.linspace(0,1,1001)
initial = theta_init[0], theta_init[1], dtheta_init[0], dtheta_init[1]
x = odeint(joint_angles, initial, t, args = (torque_array, B))
I get the error that I cannot index into torques using the t array, which makes perfect sense, however I am not sure how to have it use the current value of the torques at each time step.
I also tried putting odeint command in a for loop and only evaluating it at one time step at a time, using the solution of the function as the initial conditions for the next loop, however the function simply returned the initial conditions, meaning every loop was identical. This leads me to suspect I've made a mistake in my implementation of the standard form but I can't work out what it is. It would be preferable however to not have to call the odeint solver in a for loop every time, and rather do it all as one.
If helpful, my initial conditions and constant values are:
theta_init = np.array([10*np.pi/180, 143.54*np.pi/180])
dtheta_init = np.array([0, 0])
L_1 = 0.3
L_2 = 0.33
I_1 = 0.025
I_2 = 0.045
M_1 = 1.4
M_2 = 1.0
D_2 = 0.16
a_1 = I_1+I_2+M_2*(L_1**2)
a_2 = M_2*L_1*D_2
a_3 = I_2
Thanks for helping!
The solver uses an internal stepping that is problem adapted. The given time list is a list of points where the internal solution gets interpolated for output samples. The internal and external time lists are in no way related, the internal list only depends on the given tolerances.
There is no actual natural relation between array indices and sample times.
The translation of a given time into an index and construction of a sample value from the surrounding table entries is called interpolation (by a piecewise polynomial function).
Torque as a physical phenomenon is at least continuous, a piecewise linear interpolation is the easiest way to transform the given function value table into an actual continuous function. Of course one also needs the time array.
So use numpy.interp1d or the more advanced routines of scipy.interpolate to define the torque function that can be evaluated at arbitrary times as demanded by the solver and its integration method.
Some background...
I am currently building a macro that will estimate the cost of an injection molding tool. These tools have cavities which are filled with plastic. The number of cavities a tool has is the number of parts that will be formed.
So far my program will determine the minimum number of cavities a tool can have based on customer demand. This number is always even. The tool should have an even number of cavities. Given the bounding length and width of a cavity, and setting a limit to how much space the cavities can occupy within the tool, I need my program to calculate the combination of number of cavities along the length and width whose difference is minimized and whose product is equal to the total number of minimum cavities the tool should have.
I am programming my macro is SolidWorks VBA. I first constructed this problem in Excel and used the solver tool. But, I am unable to find a way to reference the Excel Solver Tool in SolidWorks to automate this optimization problem. I am hoping to find a clever set of equations that can solve this specific problem for me. But if someone else has a better idea of what to use, that would be awesome.
Rephrasing in an optimization format...
Variables
x = number of cavities along width of tool
y = number of cavities along length of tool
z = suggested number of total cavities
Objective Function
Minimize x - y
Such that
x * y = z
x >= 1
y >= 1
x <= y
x is an integer
y is an integer
Example
My macro says that in order to meet demand, our tool needs to have at least 48 cavities. Find the number of cavities along the length and width of the tool such that the difference is minimized and the product is equal to 48. Ideally in this case the macro would return x = 6 and y = 8.
Thanks!
Just to clarify, in the question did you actually mean to Min y-x rather than Min x-y? Otherwise there is a naïve solution taking x = 1 and y = z. Min x - y = 1-z.
I don't program in VBA but here is the idea.
Since x and y are positive integers and there product is z, with x <= y. You can essentially start with x = floor(sqrt(z)) and decrement until x = 1.
For each x, check if there exists an integer y such that x * y = z. If there is, break the loop and that's the pair you are looking for. Otherwise continue until x = 1
If you need any pseudo code so you can translate it into VBA. Here it is
int x, y;
for (x = floor(sqrt(z)); x >= 1; --x)
{
y = z / x;
if (x * y == z)
break;
}
I think you can just test out a few examples. No fancy algorithm is needed.
If you relax the condition to be 2 numbers, x and y, whose product is z and with a minimum difference, then the answer is SQRT(z).
That is not an integer that meets your needs (in general). However, you can then try integers around the square root to see if they divide z. The first one you hit (i.e. minimum difference from SQRT(z)) should have the minimum difference.
If you relax the condition to be |z - x * y| is minimized, then I would recommend testing the numbers around sqrt(z). You need to check two cases -- the floor and ceiling of the square root (and the appropriate other number).
Just in case someone is needs something similar to this in the future, but can't figure out the pseudo-code I went ahead wrote it up. I wasn't sure how to output it as two values so I just threw them together as a string for the user to see.
Option Explicit
Function Factors(ByVal Test As Long) As String
Dim Val As Long
Dim i As Long
Val = Test
i = Int(Sqr(Val))
While Val / i >= 2
If Int(Val / i) * i = Val Then
Factors = i & " & " & Val / i
Exit Function
End If
i = i - 1
Wend
End Function
I have some float numbers like:
586.3212341231,-847.3829941845
I want to use sigmoid function to make the floats in a range of [1,-1] for example :
0.842931342,-0.481238571
Any thoughts about it?
I tried scipy but it gives me wrong outcomes.
There are many such functions: see this Wikipedia link for examples. The graphic there in particular gives examples resulting in your desired range, though the endpoints 1 and -1 are not obtainable for finite values of the parameter x.
The simplest function there is
x / (1 + abs(x))
If you really want your first two sample float numbers to be mapped to your second two float numbers, you can tune your function to be
(a + x) / (b + abs(x))
for particular values of a and b. For two desired values of x and f(x) you can find a and b by solving two simultaneous linear equations. I used sympy to get the following for your sample values:
a = 246.362120860444
b = 401.521207478205
So your final resulting function is
(246.362120860444 + x) / (401.521207478205 + abs(x))
I tested this and it gives just the values you want. Here are two plots showing that this gives your desired range (-1, 1). The first one shows your two points best.
100e^0.25*y = 97.5
Solving for y
Using Excel Solver
I tried using empty column entry for y in 'By changing cells' and Set objective function as LHS of above equation (empty column entry in equation included) equal to value of 97.5 in solver.
It gives no solution
How do I do this?
It's a bit ambiguous what you're asking...
Literal math interpretation: 100*(e^0.25)*y = 97.5
Then y = 97.5 / ( 100 * exp(.25)) = .759
My guess of what you want: 100*e^(0.25*y) = 97.5
Then y = ln(97.5/100) / .25 = -.101
Another possibility: (100 * e)^(0.25 * y) = 97.5
Then y = (ln(97.5) / ln(100*e)) / .25 = 3.268
Whatever it is, this doesn't need solver!
You don't really need the solver. Just re-arrange your formula to solve for Y. Since y = b^x is the same as log(b)Y = x (log of Y, with base b)
Your formula above is the same as:
Y = (log(100e)97.5))/.25
(Read aloud, that's log of 97.5, with base 100e, divided by .25
So, Y = 3.268305672
(Bonus points for someone who can tell me how to format this so the Log looks correct)
The question is "How do I solve this exponential equation on Excel Solver?" which is a fair enough question, as it points to trying to understand how to set up solver.
My interpretation of the equation provided is given in this screenshot ...
The solver dialog box is then setup as follows ...
Of note:
This is a non-linear equation and needs GRG Nonlinear. If you choose LP Simplex, it will not pass the linearity test.
Ensure "Make Unconstrained Variables Non-Negative" is not checked.
It provided this result for me ...
A more precise answer can be obtained by decreasing the "Convergence" value on the GRG Non-Linear Options dialog.
A problem this simple can also be solved using Goal Seek.